我正在尝试更新mysql数据库,但每次我都收到Notice:Array to string conversion。我真的无法弄清楚我在这里做错了什么。有人可以帮忙吗?
我的表格
<div class="panel-body">
<form role="form" method="post" action="client_post.php" class="form-horizontal">
<fieldset>
<div class="form-group">
<label class="col-sm-3 control-label">Hospital No:
</label>
<div class="col-sm-5">
<input required class="form-control" name="Hospital_no" placeholder="Patients' Hospital Number">
</div>
</div>
<div class="form-group">
<label class="col-sm-3 control-label">Date of New Status</label>
<div class="form-group">
<label class="col-sm-3 control-label">New Status</label>
<div class="col-sm-5">
<select required name ="art_status" class="form-control">
<option></option>
<option value = "ART Restart" name="ART Restart">ART Restart</option>
<option value = "ART Transfer Out" name="ART Transfer Out" >ART Transfer Out</option>
<option value = "Pre ART Transfer Out" name="Pre ART Transfer Out">Pre ART Transfer Out</option>
<option value = "Lost to Followup" name="Lost to Followup">Lost to Followup</option>
<option value = "Stopped Treatment" name="Stopped Treatment">Stopped Treatment</option>
<option value = "Known Death" name="Known Death">Known Death</option>
</select>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-3 col-sm-5">
<input class="btn btn-primary" type="submit" value="Update" name="submit" >
</div>
</div>
</fieldset>
</form>
我试图从此页面进行更新(client.php)
if($_SERVER['REQUEST_METHOD'] === 'POST')
{
$Hosp_no = ['Hospital_no'];
$art_status = ['art_status'];
$art_date = ['art_start_date'];
$update_client = "UPDATE `pat_reg` SET `art_status` = '$art_status', `art_start_date` = '$art_date' WHERE `Hospital_no` = '$Hosp_no'";
if (mysqli_query($dbcon,$update_client))
{
echo"<script>alert('Record Updated')</script>";
//echo"<script>window.open('client_update.php','_self')</script>";
}
$dbcon->close();
}
请帮忙。
答案 0 :(得分:3)
使用$_POST
超全局允许您访问表单中提交的值。
更改以下内容:
$Hosp_no = ['Hospital_no'];
$art_status = ['art_status'];
$art_date = ['art_start_date'];
到此:
$Hosp_no = $_POST['Hospital_no'];
$art_status = $_POST['art_status'];
$art_date = $_POST['art_start_date'];
PS。您应该在查询数据库之前逃避用户的输入(或者,理想情况下,使用prepared statements),否则如果向公众发布,您可能会很快被黑客攻击。